What is the integral of 1/((x^2+2x)^{3/2)} ?

The integral of 1/((x^2+2x)^{3/2)} is -(x+1)/((x^2+2x)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x^2+2x)^{3/2)}

\int \frac{1}{( x ^{2} + 2 x ) ^{\frac{3}{2}}} d x

- \frac{x + 1}{( x ^{2} + 2 x ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} + 2 x ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} + 2 x : (x + 1)^{2} - 1

= \int \frac{1}{( ( x + 1 ) ^{2} - 1 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( u ^{2} - 1 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( u ^{2} - 1 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int cot^{2} (v) sec (v) d v

= \int cot^{2} (v) sec (v) d v

Simplify

cot^{2} (v) sec (v) : cot (v) csc (v)

= \int cot (v) csc (v) d v

Apply u-substitution

: \int - 1 d w

= \int - 1 d w

Integral of a constant:

\int a d x = a x

= (- 1) w

Substitute back

= (- 1) csc (arcsec (x + 1))

Simplify

(- 1) csc (arcsec (x + 1)) : - \frac{x + 1}{( x ^{2} + 2 x ) ^{\frac{1}{2}}}

= - \frac{x + 1}{( x ^{2} + 2 x ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x + 1}{( x ^{2} + 2 x ) ^{\frac{1}{2}}} + C

\int \frac{1}{16 x ^{2} - 24 x - 7} d x

\int \frac{9 csc ^{2} ( 3 x )}{3 - cot ( 3 x )} d x

\int \frac{1}{cos ( 4 θ )} d θ

\int \frac{3 x + 3}{x ^{3} - x ^{2} - 2 x} d x

\int \frac{x ^{l n (x)} ln ( x )}{x} d x

What is the integral of 1/((x^2+2x)^{3/2)} ?
The integral of 1/((x^2+2x)^{3/2)} is -(x+1)/((x^2+2x)^{1/2)}+C